Xavier Aptitude Test 2014 Paper

Question : 66 of 103

Marks: +1, -0

Aditya has a total of 18 red and blue marbles in two bags (each bag has marbles of both colors). A marble is randomly drawn from the first bag followed by another randomly drawn from the second bag, the probability of both being red is 5/16. What is the probability of both marbles being blue?

1/16
2/16
3/16
4/16
None of the above.

Validate

Solution:

(c)

	Red	Blue
Bag 1	$X_1$	$Y_1$
Bag 2	$X_2$	$y_2$

\ ( {X}_{1}+\ {Y}_{1}\ )+\ (\ {X}_{2}+\ {Y}_{2}\ )=18

............(i)
Selecting a marble from the first bag and then from thesecond bag can be done in

\ (\ {X}_{1}+\ {Y}_{1}\ ) × (\ {X}_{2}+\ {Y}_{2}\ )

ways.
Selecting a red marble from thefirst bag and then a redmarble from the second bag can be done in

\ (\ {X}_{1}\ ) ×( {X}_{2}\ )

ways.

\∴

Probability of selecting red marbles from both thebags

= (\ {X}_{1}\ ) ×(\ {X}_{2} ) \/ (\ {X}_{1}+\ {Y}_{1} ) \×(\ {X}_{2}+\ {Y}_{2}\ )=5 / 16

Let

\ (\ {X}_{1}+\ {Y}_{1} ) \×(\ {X}_{2}+\ {Y}_{2} )=16 {a} \ ........

(ii)

\⋅(\ {X}_{1}\ ) ×(\ {X}_{2}\ )=5 {a} \ ............

(iii)
Considering(i) and (ii), a = 2 or 5
Case I : a = 2

\ (\ {X}_{1}+\ {Y}_{1} ) ×(\ {X}_{2}+\ {Y}_{2}\ )=32\ .........

(iv)
∴ From (i) and (iv).

\ (\ {X}_{1}+\ {Y}_{1}\ )=2

and

\ (\ {X}_{2}+\ {Y}_{2}\ )=16

\∴{X}_{1}=\ {Y}_{1}=1 \⇒ \ {X}_{2}=10\ (...(\ {X}_{1} ) ×( {X}_{2}\ )=10 )

\∴{Y}_{2}=6

Probability of both marbles being blue = (1 × 6)/32
= 3 /16
Case II : a = 5

(\ {X}_{1}+ {Y}_{1}\ ) \×(\ {X}_{2} \× \ {Y}_{2}\)=80 \ \........( {v})

∴ From (i) and (v),

( {X}_{1}+\ {Y}_{1}\ )=8

and

( {X}_{2}+\ {Y}_{2}\ )=10

\(\∵ (\ {X}_{1} ) ×(\ {X}_{2}\ )=25\ )

\ {Y}_{1}=3

and

{Y}_{2}=5

Probability of both marbles being blue

=(5 \× 3) / 80=3/ 16

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